Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T nx(T x) {
  return ++x;
}
template <class T>
inline T pr(T x) {
  return --x;
}
template <class T>
inline void get(T &n) {
  char c = getchar();
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  n = 0;
  T s = 1;
  if (c == '-') s = -1, c = getchar();
  while (c >= '0' && c <= '9') n *= 10, n += c - '0', c = getchar();
  n *= s;
}
const int maxn = 10010;
int n;
int a[maxn], b[maxn], bt[maxn], f[40], g[40];
int isf[maxn], isg[maxn];
vector<pair<int, int> > ans, bns;
void XOR(int i, int j) {
  a[i] ^= a[j];
  ans.push_back(pair<int, int>(i, j));
}
void YOR(int i, int j) {
  b[i] ^= b[j];
  bns.push_back(pair<int, int>(i, j));
}
int main() {
  int i, j, k, t, tt, x, y;
  scanf("%d", &n);
  for (i = (1); i <= (n); ++i) scanf("%d", a + i);
  for (i = (1); i <= (n); ++i) scanf("%d", b + i);
  for (i = (1); i <= (n); ++i) {
    for (x = (30); x >= (1); --x)
      if (((a[i]) & (1LL << ((x)-1)))) {
        if (f[x])
          XOR(i, f[x]);
        else {
          f[x] = i;
          isf[i] = x;
          break;
        }
      }
  }
  for (i = (1); i <= (n); ++i) {
    for (x = (30); x >= (1); --x)
      if (((b[i]) & (1LL << ((x)-1)))) {
        if (g[x])
          YOR(i, g[x]);
        else {
          g[x] = i;
          isg[i] = x;
          break;
        }
      }
  }
  for (x = (30); x >= (1); --x)
    if (g[x] && g[x] != f[x]) {
      i = f[x];
      j = g[x];
      XOR(i, j);
      XOR(j, i);
      XOR(i, j);
      swap(isf[i], isf[j]);
      f[isf[i]] = i;
      f[isf[j]] = j;
    }
  for (x = (30); x >= (1); --x)
    if (g[x]) {
      i = g[x];
      for (y = (x - 1); y >= (1); --y)
        if (((a[i]) & (1LL << ((y)-1))) != ((b[i]) & (1LL << ((y)-1))))
          XOR(i, f[y]);
      if (a[i] != b[i]) break;
    }
  if (x >= 1) {
    puts("-1");
    return 0;
  }
  for (i = (1); i <= (n); ++i)
    if (!isg[i]) XOR(i, i);
  for (i = (int((bns).size()) - 1); i >= (0); --i) ans.push_back(bns[i]);
  printf("%d\n", int((ans).size()));
  for (i = (0); i <= (int((ans).size()) - 1); ++i)
    printf("%d %d\n", ans[i].first, ans[i].second);
  return 0;
}